MECHANICS II – DYNAMICS
(BOOK I)
Dynamics - the part of mechanics that deals with the analysis of bodies in motion.
Kinematics - the study of the geometry of motion. Kinematics is used to relate displacement,
velocity, acceleration, and time, without reference to the cause of the motion.
Kinetics - the study of the relation between the forces acting on a body, the mass of the body, and
the motion of the body. Kinetics is used to predict the motion caused by given forces or to
determine the forces required to produce a given motion.
Rectilinear motion - A particle moving along a straight line
The average velocity (over the time interval Δt) - the quotient of the displacement Δx and the time
interval Δt
Average Velocity =
∆𝑥
∆𝑡
The instantaneous velocity v - obtained from the average velocity by choosing shorter and shorter time
intervals Δt and displacements Δx
∆𝑥 1
Instantaneous velocity = v = lim ( )
∆𝑡→0
∆𝑡
Observing that the limit of the quotient is equal, by definition, to the derivative of x with respect
to t, we write;
𝑣=
𝑑𝑥
𝑑𝑡
The average acceleration - quotient of Δv and Δt:
Average acceleration =
∆𝑣
∆𝑡
The instantaneous acceleration a - obtained from the average acceleration by choosing smaller and
smaller values for Δt and Δv
Instantaneous acceleration = a = lim
∆𝑣
∆𝑡 →0 ∆𝑡
=
𝑑𝑣
𝑑𝑡
=
𝑑2𝑥
𝑑𝑡 2
Deceleration - refers to “a” when the speed (the magnitude of v) decreases
=𝑣
𝑑𝑣
𝑑𝑥
DETERMINATION OF THE MOTION OF A PARTICLE
The motion of a particle is said to be known if the position of the particle is known for every
value of the time t.
More often, the conditions of the motion will be specified by the type of acceleration that the
particle possesses.
the acceleration of the particle can be expressed as a function of one or more of the
variables x, v, and t.
Three common classes of motion
1) The Acceleration Is a Given Function of time - a = f(t)
dv = a dt
dv = f(t) dt
∫ 𝑑𝑣 = ∫ 𝑓(𝑡)
In order to uniquely define the motion of the particle, it is necessary to specify the initial conditions of the
motion
𝑣
𝑡
𝑡
∫𝑣 𝑑𝑣 = ∫0 𝑓(𝑡)𝑑𝑡
𝑣 − 𝑣0 = ∫0 𝑓(𝑡)𝑑𝑡
0
2) The Acceleration Is a Given Function of position – a = f(x)
𝑎=𝑣
𝑑𝑣
𝑣 𝑑𝑣 = 𝑎 𝑑𝑥
𝑑𝑥
𝑣
𝑥
0
0
𝑣 𝑑𝑣 = 𝑓(𝑥)𝑑𝑥
𝑥
1 2
1 2
𝑣 − 𝑣 = ∫ 𝑓(𝑥)𝑑𝑥
𝑥0
2
2 0
∫𝑣 𝑣 𝑑𝑣 = ∫𝑥 𝑓(𝑥)𝑑𝑥
3) The Acceleration Is a Given Function of velocity – a = f(v)
𝑓(𝑣) =
𝑑𝑣
𝑑𝑡
𝑑𝑡 =
𝑑𝑥
𝑣
𝑓(𝑣) = 𝑣
𝑑𝑣
𝑑𝑥
UNIFORMLY ACCELERATED RECTILINEAR MOTION
Uniformly accelerated rectilinear motion - the acceleration a of the particle is constant
Where a = dv / dt = constant
The velocity v of the particle is obtained by integrating this equation:
𝑣
𝑡
∫𝑣 𝑑𝑣 = 𝑎 ∫0 𝑑𝑡
𝑣 − 𝑣0 = 𝑎𝑡
0
𝑣 = 𝑣0 + 𝑎𝑡
Substituting for velocity;
𝑑𝑥
𝑑𝑡
𝑥
= 𝑣0 + 𝑎𝑡
𝑡
∫𝑥 𝑑𝑥 = ∫0 (𝑣0 + 𝑎𝑡) 𝑑𝑡
0
1
1
𝑥 − 𝑥0 = 𝑣0 𝑡 + 𝑎𝑡 2
𝑥 = 𝑥0 + 𝑎𝑡 2
2
2
Another equation for velocity without time-dependency could be derived as;
𝑎=
𝑑𝑣
𝑑𝑡
, 𝑎=𝑣
𝑑𝑣
𝑑𝑥
,
𝑥
𝑣
0
0
𝑎 ∫𝑥 𝑑𝑥 = ∫𝑣 𝑣 𝑑𝑣 ,
1
𝑎(𝑥 − 𝑥0 ) = (𝑣 2 − 𝑣𝑜2 ), 𝑣 2 = 𝑣02 + 2𝑎(𝑥 − 𝑥0 )
2
It is important to keep in mind that the three equations above can be used only when the
acceleration of the particle is known to be constant.
MOTION OF SEVERAL PARTICLES
**When several particles move independently along the same line, independent equations of motion can be
written for each particle.
**Time should be recorded from the same initial instant for all particles, and displacements should be
measured from the same origin and in the same direction.
Relative Motion of Two Particles
Relative position coordinate - two particles A and B moving along the same straight line xB -xA defines
the relative position coordinate of B with respect to A and is denoted by xB/A .
xB/A = xB -xA
xB =xA +xB/A
or
Regardless of the positions of A and B with respect to the origin, a positive sign for xB/A means that B is to
the right of A, and a negative sign means that B is to the left of A.
Relative velocity - The rate of change of xB/A (for B)
vB/A = vB -vA
or
vB =vA +vB/A
Relative acceleration - The rate of change of vB/A (for B)
aB/A = aB - aA
or
aB =aA + aB/A
GRAPHICAL SOLUTION OF RECTILINEAR- MOTION PROBLEMS
the area measured under the v−t curve from t1 to t2 is equal to the change in x during that time
interval.
the area measured under the a–t curve from t1 to t2 is equal to the change in v during that time
interval.
In general, if the acceleration is a polynomial of degree n in t, the velocity will be a polynomial of
degree n+1 and the position coordinate a polynomial of degree n+2; these polynomials are
represented by motion curves of a corresponding degree.
moment-area method- allows us to calculate 𝑥1 − 𝑥0 directly from a-t curve (x and v at t = 0 as x0 and v0
and their values at t = t1 as x1 and v1 )
𝑣1
𝑥1 − 𝑥0 = 𝑣0 𝑡1 + ∫ (𝑡1 − 𝑡) 𝑑𝑣
𝑣2
(BOOK II)
CURVILINIEAR MOTION OF PARTICLES
Derivatives of Vector Functions
𝑑(𝑃 + 𝑄)
∆(𝑃 + 𝑄)
∆𝑃 ∆𝑄
𝑑𝑃 𝑑𝑄
= lim
= lim ( +
)=
+
∆𝑢→0
∆𝑢→0 ∆𝑢
𝑑𝑢
∆𝑢
∆𝑢
𝑑𝑢 𝑑𝑢
(𝑓 + ∆𝑓)(𝑃 + ∆𝑃) − 𝑓𝑃
𝑑(𝑓𝑃)
∆𝑓
∆𝑃
= lim
= lim ( 𝑃 + 𝑓 ( ))
∆𝑢→0
∆𝑢→0 ∆𝑢
𝑑𝑢
∆𝑢
∆𝑢
Scalar Product;
𝑑(𝑃 ∗ 𝑄) 𝑑𝑃
𝑑𝑄
=
𝑄+𝑃
𝑑𝑢
𝑑𝑢
𝑑𝑢
Vector Product;
𝑑(𝑃𝑥𝑄) 𝑑𝑃
𝑑𝑄
=
𝑥𝑄+𝑃𝑥
𝑑𝑢
𝑑𝑢
𝑑𝑢
Motion Relative to a Frame In Translation
𝑟𝐵 = 𝑟𝐴 + 𝑟𝐵/𝐴
𝑣𝐵 = 𝑣𝐴 + 𝑣𝐵/𝐴
𝑎𝐵 = 𝑎𝐴 + 𝑎𝐵/𝐴
TANGENTIAL AND NORMAL COMPONENTS
Plane Motion of a Particle
Assume a particle that is on point P at t, and on P’ at t + Δt;
𝑒𝑡 is the tangent unit vector drawn from P
𝑒𝑡 ′ is the tangent unit vector drawn from P’
When 𝑒𝑡 and 𝑒𝑡 ′ drawn from the same origin; which would make them the radius lengths of 1 unit.
The angle between being Δ𝜃
The magnitude Δ𝑒𝑡 would equal = 2 sin (Δ𝜃 / 2)
Then, the unit vector along the normal to the path of the particle 𝒆𝒏 .
lim
Δ𝜃→0
2 sin(
Δ𝜃
)
2
Δ𝜃
=1
𝑒𝑛 =
𝑑𝑒𝑡
𝑑𝜃
Now, using this information to rewrite some already known equations;
V = v𝑒𝑡
a= dv/dt = (dv/dt) 𝑒𝑡 + v (d𝑒𝑡 /dt)
d𝑒𝑡 /dt = (d𝑒𝑡 / d 𝜃) * (d 𝜃/ds) * (ds/dt)
ds/dt = v
d𝑒𝑡 / d 𝜃 = 𝒆𝒏
d 𝜃/ds = 1/ 𝜌
𝝆 is the radius of the curvature of the path (at P)
𝑑𝑒𝑡 𝑣
= 𝑒𝑛
𝑑𝑡
𝜌
Acceleration in Curvilinear Plane Motion
tangential and normal components - Cartesian Coordinates
𝑑𝑣
𝑣2
𝑎 =
𝑒 +
𝑒𝑛
𝑑𝑡 𝑡
𝜌
(acceleration = tangential + normal (centrifugal))
radial and transverse components ~ Polar Coordinates
The vector 𝑒𝑟 is directed along the line of action; and the vector 𝑒𝜃 is obtained by rotating 𝑒𝑟 through 90°
counterclockwise – normal to the line of action.
𝑑𝑒𝑟
𝑑𝜃
𝑎𝑛𝑑
− 𝑒𝑟 =
𝑑𝜃
𝑑𝑒𝑟 𝑑𝜃
𝑑𝑒𝑟
=
=
𝑑𝑡
𝑑𝜃 𝑑𝑡
𝑑𝑡
𝑎𝑛𝑑
− 𝑒𝑟
𝑒𝑟̇ = 𝜃̇ 𝑒𝜃
𝑎𝑛𝑑
𝑒𝜃̇ = −𝜃̇ 𝑒𝑟
𝑒𝜃 =
𝑑𝑒𝜃
𝑑𝜃
Rewritten with the chain rule;
𝑒𝜃
𝑑𝜃
𝑑𝑒𝜃 𝑑𝜃
𝑑𝑒𝜃
=
=
𝑑𝑡
𝑑𝜃 𝑑𝑡
𝑑𝑡
Finalising;
Using newly obtained polar axis unit vectors;
Velocity relation;
𝑣 = 𝑟̇ 𝑒𝑟 + 𝑟𝜃̇ 𝑒𝜃
Acceleration relation;
𝑎 = (𝑟̈ − 𝑟𝜃̇ 2 )𝑒𝑟 + (𝑟𝜃̈ + 2𝑟̇ 𝜃̇ )𝑒𝜃
Kinetics of Particles: Newton’s Second Law
NEWTON’S SECOND LAW OF MOTION
If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to
the magnitude of the resultant and in the direction of this resultant force.
LINEAR MOMENTUM OF A PARTICLE
Recalling the Newton’s Second Law;
∑𝐹 =
𝑑
(𝑚𝑣)
𝑑𝑡
Where mv is the linear momentum;
𝐿 = 𝑚𝑣
Which these equations show that; “the resultant of the forces acting on the particle is equal to the rate of
change of the linear momentum of the particle. “
∑ 𝐹 = 𝐿̇
conservation of linear momentum - if the resultant force acting on a particle is zero, the linear
momentum of the particle remains constant, in both magnitude and direction.
EQUATIONS OF MOTION
Rectangular Components
Fx = m𝒂𝒙 = 𝒎𝒙̈ Fy = m𝒂𝒚 = 𝒎𝒚̈
Fz = m𝒂𝒛 = 𝒎𝒛̈
Tangential and Normal Components
𝜮𝑭𝒕 = 𝒎𝒂𝒕 = 𝒎
𝒅𝒗
𝒅𝒕
DYNAMIC EQUILIBRIUM
𝜮𝑭𝒏 = 𝒎𝒂𝒏 = 𝒎
𝒗𝟐
𝝆
dynamic equilibrium - if we add the vector -ma to the forces acting on the particle, we obtain a system of
vectors equivalent to zero;
∑ 𝐹 − 𝑚𝑎 = 0
Inertia Vector - The vector -ma, of magnitude ma and of direction opposite to that of the acceleration.
(Book III)
Angular Momentum
angular momentum - The moment about origin of the vector mv / moment of momentum (𝐻𝑂 )
𝐻𝑜 = 𝑟 𝑥 𝐿 = 𝑟 𝒙 𝑚𝑣 = 𝑟𝑚𝑣 𝑠𝑖𝑛𝜙
If polar coordinates are used, we resolve the linear momentum of the particle into radial and transverse
components;
𝐻𝑜 = 𝑟𝑚𝑣𝑠𝑖𝑛𝜙 = 𝑟𝑚𝑣𝜃
Knowing that, 𝑣𝜃 = 𝑟𝜃̇
𝐻𝑜 = 𝑚𝑟 2 𝜃̇
If the r x mv equation is differentiated; we are left with “r x ma” which is r x F representing the sum of
moments;
The sum of the moments about O of the forces acting on the particle is equal to the rate of change of
the moment of momen- tum, or angular momentum, of the particle about O.
∑ 𝑀𝑂 = 𝐻𝑂̇
Newton’s Law of Gravitation
𝐹=𝐺
𝑔=
𝑀𝑚
𝑟2
𝐺𝑀
𝑅2
Trajectory of a Particle Under a Central Force
Application to Space Mechanics
Kinetics of Particles: Energy and Momentum Methods
WORK OF A FORCE
Work (Nm)(Joule)
𝑑𝑈 = 𝐹 𝑑𝑟 = 𝐹 𝑑𝑠 𝑐𝑜𝑠𝛼
𝐴2
𝑈1→2 = ∫ 𝐹 𝑑𝑟
𝑎1
Combining these two equations;
𝑆2
𝑆2
𝑈1→2 = ∫ (𝐹 𝑐𝑜𝑠𝛼) 𝑑𝑠 = ∫ 𝐹𝑡 𝑑𝑠
𝑆1
𝑆1
Work of a Constant Force in Rectilinear Motion
𝑈1→2 = ( 𝐹 𝑐𝑜𝑠𝛼) ∆𝑥
Work of the Force of Gravity
𝑦2
𝑈1→2 = − ∫ 𝑊 𝑑𝑦 = 𝑊(𝑦2 − 𝑦1 ) = −𝑊 Δ𝑦
𝑦1
Work of the Force Exerted by a Spring
𝑥2
1
1
𝑈1→2 = − ∫ 𝑘𝑥 𝑑𝑥 = 𝑘𝑥12 − 𝑘𝑥22
2
2
𝑥1
𝑜𝑟
1
𝑈1→2 = − (𝐹1 + 𝐹2 ) ∆𝑥
2
Work of a Gravitational Force
𝑟2
𝐺𝑀𝑚
𝐺𝑀𝑚 𝐺𝑀𝑚
𝑑𝑟 =
−
2
𝑟
𝑟2
𝑟1
𝑟1
𝑈1→2 = − ∫
KINETIC ENERGY OF A PARTICLE
POWER AND EFFICIENCY
mechanical efficiency
𝜂=
POTENTIAL ENERGY
𝑃𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡
𝑃𝑜𝑤𝑒𝑟 𝐼𝑛𝑝𝑢𝑡
Conservation of energy
Motion Under a Conservative Central Force
Principle of Impulse and momentum
Using the relation of force and linear momentum, and differentiating both sides then, integration from
one instant of time to another; we would derive an equation as;
𝑡2
∫ 𝐹 𝑑𝑡 = 𝑚𝑣2 − 𝑚𝑣1
𝑡1
Rearranging the last term by transposing;
𝑡2
𝑚𝑣1 + ∫ 𝐹 𝑑𝑡 = 𝑚𝑣2
𝑡1
the final momentum m𝑣2 of the particle can be obtained by adding vectorially its initial momentum
m𝑣1 and the impulse of the force F during the time interval considered.
linear impulse - force F during the interval of time considered
𝑡2
𝐼𝑚𝑝1→2 = ∫ 𝐹 𝑑𝑡 = 𝐹(𝑡2 − 𝑡1 ) 𝑤ℎ𝑒𝑛 𝐹 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡
𝑡1
Finalising as;
𝑚𝑣1 + 𝐼𝑚𝑝1→2 = 𝑚𝑣2
When several forces act on a particle;
𝑚𝑣1 + Σ𝐼𝑚𝑝1→2 = 𝑚𝑣2
When a problem involves two particles or more, each particle can be considered separately and can be
written for each particle.
We can also add vectorially the momenta of all the particles and the impulses of all the forces involved.
Σ𝑚𝑣1 + Σ𝐼𝑚𝑝1→2 = Σ𝑚𝑣2
IMPULSIVE MOTION
Impulsive force - A force acting on a particle during a very short time interval that is large enough to
produce a definite change in momentum.
𝑚𝑣1 + Σ𝐹 ∆𝑡 = m𝑣2
Nonimpulsive forces - the weight of the body, the force exerted by a spring, or any other force which is
known to be small compared with an impulsive force. ~ neglegted.
impulsive motion of several particles;
Σ𝑚𝑣1 + Σ𝐹 ∆𝑡 = Σ𝑚𝑣2
(Book IV)
Impact
A collision between two bodies which occurs in a very small interval of time and during which the two
bodies exert relatively large forces on each other.
line of impact - The common normal to the sur- faces in contact during the impact
central impact – where the mass centers on the two colliding bodies are located on the line of
impact
eccentric impact – when mass centers do not corrolate upon the line of impact.
oblique impact – when the velocities of the two colliding particles are not directed along the line of
impact.
DIRECT CENTRAL IMPACT
Consider two particles A and B, which are moving in the same straight line and to the right.
If vA is larger than vB, particle A will eventually strike particle B.
Under the impact, the two particles, will deform and, at the end of the period of deformation, they will
have the same velocity u.
A period of restitution will then take place, the two particles either will have regained their original shape
or will stay permanently deformed.
Conservation of momentum;
𝑚𝐴 𝑣𝐴 + 𝑚𝐵 𝑣𝐵 = 𝑚𝐴 𝑣𝐴′ + 𝑚𝐵 𝑣𝐵 ′
Assuming the force on A exerted by B is P and the force on B exerted by A is R;
𝑚𝐴 𝑣𝐴 − ∫ 𝑃 𝑑𝑡 = 𝑚𝐴 𝑢
𝑚𝐴 𝑢 − ∫ 𝑅 𝑑𝑡 = 𝑚𝐴 𝑣𝐴 ′
coefficient of restitution - The ratio of the magnitudes of the impulses corresponding to the period of
restitution and to the period of deformation;
𝑒=
𝑣𝐵′ − 𝑣𝐴 ′
∫ 𝑅 𝑑𝑡 𝑢 − 𝑣𝐴′ 𝑣𝐵′ − 𝑢
=
=
=
𝑣𝐴 − 𝑣𝐵
∫ 𝑃 𝑑𝑡 𝑣𝐴 − 𝑢 𝑢 − 𝑣𝐵
𝑣𝐵′ − 𝑣𝐴′ = 𝑒(𝑣𝐴 − 𝑣𝐵 )
Oblique Central Impact
Systems of Particles
internal force - The force exerted on 𝑃𝑖 by another particle Pj of the system / 𝑓𝑖𝑗
The resultant of the internal forces;
𝑛
∑ 𝑓𝑖𝑗
𝑗=1
The resultant of all the external forces;
𝑛
𝐹𝑖 + ∑ 𝑓𝑖𝑗 = 𝑚𝑖 𝑎𝑖
𝑗=1
Taking the moments;
𝑛
𝑟𝑖 𝑥 𝐹𝑖 + ∑(𝑟𝑖 𝑥 𝑓𝑖𝑗 ) = 𝑟𝑖 𝑥 𝑚𝑖 𝑎𝑖
𝑗=1
effective forces - The vectors 𝑚𝑖 𝑎𝑖 are referred.
Newton’s law of gravitation to particles acting at a distance;
the forces fij and fji are equal and opposite and have the same line of action;
𝑓𝑖𝑗 + 𝑓𝑗𝑖 = 0
𝑟𝑖 𝑥 𝑓𝑖𝑗 + 𝑟𝑗 𝑥 𝑓𝑗𝑖 = 𝑟𝑖 𝑥 (𝑓𝑖𝑗 + 𝑓𝑗𝑖 ) + (𝑟𝑗 − 𝑟𝑖 ) 𝑥 𝑓𝑗𝑖 = 0
(𝑟𝑗 − 𝑟𝑖 ) 𝑎𝑛𝑑 𝑓𝑗𝑖 are collinear
𝑛
𝑛
𝑛
𝑛
∑ ∑ 𝑓𝑖𝑗 = 0 , ∑ ∑(𝑟𝑖 𝑥 𝑓𝑖𝑗 ) = 0
𝑖=1 𝑗=1
𝑖=1 𝑗=1
the system of the external forces acting on the particles and the system of the effective forces of the particles
are equipollent.
LINEAR AND ANGULAR MOMENTUM OF A SYSTEM OF PARTICLES
∑ 𝐹 = 𝐿̇
∑ 𝑀0 = 𝐻𝑜̇
MOTION OF THE MASS CENTER OF A SYSTEM OF PARTICLES
𝑛
𝑚𝑟⃗ = ∑ 𝑚𝑖 𝑟𝑖
𝑖=1
Differentiating the equation once and twice;
𝐿 = 𝑚𝑣⃗
𝐿̇ = 𝑚𝑎⃗
∑ 𝐹 = 𝑚𝑎⃗
the mass center of a system of particles moves as if the entire mass of the system and all the external forces
were concentrated at that point.
ANGULAR MOMENTUM OF A SYSTEM OF PARTICLES ABOUT ITS MASS CENTER
consider the motion of the particles of the system with respect to a centroidal frame of reference Gx’y’z’
𝑛
𝐻𝐺′ =
∑(𝑟𝑖′ 𝑥 𝑚𝑖 𝑣𝑖′ )
𝑖=1
𝑛
𝐻𝐺̇′ =
∑(𝑟𝑖′ 𝑥 𝑚𝑖 𝑎𝑖′ )
𝑖=1
For acceleration;
𝑎𝑖 = 𝑎⃗ + 𝑎𝑖′
𝑛
𝐻𝐺̇′ =
𝑛
∑(𝑟𝑖′
𝑥 𝑚𝑖 𝑎𝑖 ) − (∑ 𝑚𝑖 𝑟𝑖′ ) 𝑥 𝑎⃗
𝑖=1
𝑖=1
Same for velocity;
𝑣𝑖 = 𝑣⃗ + 𝑣𝑖′
𝑛
𝐻𝐺̇′ =
𝑛
(∑ 𝑚𝑖 𝑟𝑖′ ) 𝑥 𝑣⃗ + ∑(𝑟𝑖′
𝑖=1
𝑖=1
Therefore;
∑ 𝑀𝐺 = 𝐻𝐺̇
𝑥 𝑚𝑖 𝑣𝑖′ )
(BOOK VI)
Kinematics of Rigid Bodies
Translation
A motion is said to be a translation if any straight line inside the body keeps the same direction during the
motion.
It can also be observed that in a translation all the particles forming the body move along parallel paths.
rectilinear translation - If these paths are straight lines
curvilinear translation – If the paths are curved lines
Rotation about a Fixed Axis
the particles forming the rigid body move in parallel planes along circles centered on the same fixed axis
~all its particles moving along concentric circles
If the axis of rotation, intersects the rigid body, the particles located on the axis have zero velocity and zero
acceleration.
General Plane Motion
Motions in which all the particles of the body move in parallel planes.
Any plane motion which is neither a rotation nor a translation is referred to as a general plane motion.
Motion about a Fixed Point - The three-dimensional motion of a rigid body attached at a fixed point O.
General Motion - Any motion of a rigid body which does not fall in any of the categories above.
Translation
let A and B be any two of a rigid body’s particles in translation;
Since 𝑟𝐵/𝐴 must always be a constant in this type of motion; its derivative will be zero.
Therefore, differentiating the equation -> 𝑟𝐵 = 𝑟𝐴 + 𝑟𝐵/𝐴 ;
𝑣𝐵 = 𝑣𝐴
𝑎𝑛𝑑
𝑎𝐴 = 𝑎𝐵
ROTATION ABOUT A FIXED AXIS
Consider a rigid body which rotates about a fixed axis AA’.
Let P be a point of the body and r its position vector with respect to a fixed frame of reference.
Assume that the frame is centered at origin on AA’ and that the z axis coincides with AA’
Let B be the projection of P on AA’
𝑣=𝑤𝑥𝑟
𝑤
⃗⃗⃗ = 𝑤𝑘 = 𝜃̇ 𝑘
𝜃 : the angular coordinate of the body
𝑎 = 𝛼 𝑥 𝑟 + 𝑤 𝑥 (𝑤 𝑥 𝑟)
𝛼 = 𝛼𝑘 = 𝑤̇ 𝑘 = 𝜃̈𝑘
Rotation of a Representative Slab
𝑣 = 𝑤𝑘 𝑥 𝑟
|𝑣| = 𝑟𝑤
𝑎 = 𝛼𝑘 𝑥 𝑟 − 𝑤 2 𝑟
𝑎𝑡 = 𝛼𝑘 𝑥 𝑟
𝑎𝑛𝑑
𝑎𝑛 = 𝑟𝑤 2
EQUATIONS DEFINING THE ROTATION OF A RIGID BODY ABOUT A FIXED AXIS
𝑤=
𝑑𝜃
𝑑𝑡
𝛼=
𝑑𝑤 𝑑 2 𝜃
𝑑𝑤
=
=𝑤
𝑑𝑡
𝑑𝑡
𝑑𝜃
Uniform Rotation - the angular acceleration is zero (𝛼 = 0);
𝜃 = 𝜃0 + 𝑤𝑡
Uniformly Accelerated Rotation - the angular acceleration is constant (𝛼 = 𝑐𝑜𝑛𝑠𝑡);
𝑤 = 𝑤0 + 𝛼𝑡
1
𝜃 = 𝜃0 + 𝑤0 𝑡 + 𝛼𝑡 2
2
𝑤 2 = 𝑤02 + 2𝛼(𝜃 − 𝜃0 )
General Plane Motion
A general plane motion can always be considered as the sum of a translation and a rotation.
ABSOLUTE AND RELATIVE VELOCITY IN PLANE MOTION
INSTANTANEOUS CENTER OF ROTATION IN PLANE MOTION